COPY with dynamic file name

Question

I am trying to write a function to load csv data into a table. I want the input argument to be the path to the file.

CREATE OR REPLACE FUNCTION public.loadd         


        

Answer 1

You need dynamic SQL:

CREATE OR REPLACE FUNCTION loaddata(filepathname text)
  RETURNS void AS
$func$
BEGIN
   EXECUTE format ('
   COPY climatedata(
         climatestationid
       , date
         ... -- more columns 
       , tminsflag)
   FROM %L (FORMAT CSV, HEADER)'  -- current syntax
           -- WITH CSV HEADER'    -- tolerated legacy syntax
   , $1);  -- pass function parameter filepathname to format() 
END
$func$ LANGUAGE plpgsql;

format() requires PostgreSQL 9.1+.
Pass the file name without extra set of (escaped) single quotes:

SELECT loaddata('/absolute/path/to/my/file.csv')

format() with %L escapes the file name safely. Would be susceptible to SQL injection without it.

---

Aside, you have a function name mismatch:

CREATE OR REPLACE FUNCTION public.loaddata(filepathname varchar)
...
ALTER FUNCTION public.filltmaxa(character varying)